+67 Find constants A and B such that
\(\frac{x+7}{x^2 -x-2}=\frac{A}{x-2} + \frac{B}{x+1}\)
for all x such that \(x \ne -1\) and \(x \ne 2\). Give your answer as the ordered pair (A,B).
Thank you
If you let x go infinity, then you get 0 = A + B.
If you let x = 0, then you get -7/2 = A/(-2) + B.
The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).
Note that x^2-x-2 factors to x-2 * x+1
Multiply the equation by (x-2)(x+1)
x+7 = a(x+1) + b (x-2)
x+7 = ax + a + bx - 2b equat the 'x' and 'non-x' components to get the following two equations
x = x (a+b) and 7 = a - 2b
1 =( a+b )
a = 1-b sub this into the red equation 7 = 1-b - 2b to find b = -2 then a = 3
